QuestionJuly 3, 2025

Calculate the Retract speed of a cylinder,given the flow rate and piston diameter. R=4.75lpm D_(p)=6.25cm D_(r)=4.75cm 366.51cm/min 379.84cm/min 381.25cm/min

Calculate the Retract speed of a cylinder,given the flow rate and piston diameter. R=4.75lpm D_(p)=6.25cm D_(r)=4.75cm 366.51cm/min 379.84cm/min 381.25cm/min
Calculate the Retract speed of a cylinder,given the flow rate and piston diameter. R=4.75lpm D_(p)=6.25cm D_(r)=4.75cm 366.51cm/min 379.84cm/min 381.25cm/min

Solution
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Answer

379.84 \, \text{cm/min} Explanation 1. Convert flow rate to cm³/min Given R = 4.75 \, \text{lpm}, convert to \text{cm}^3/\text{min}: R = 4.75 \times 1000 = 4750 \, \text{cm}^3/\text{min}. 2. Calculate piston area Piston diameter D_p = 6.25 \, \text{cm}. Area A_p = \pi \left(\frac{D_p}{2}\right)^2 = \pi \left(\frac{6.25}{2}\right)^2. 3. Calculate rod area Rod diameter D_r = 4.75 \, \text{cm}. Area A_r = \pi \left(\frac{D_r}{2}\right)^2 = \pi \left(\frac{4.75}{2}\right)^2. 4. Calculate effective area for retraction Effective area A_e = A_p - A_r. 5. Calculate retract speed **Formula**: V = \frac{R}{A_e}. Substitute R = 4750 \, \text{cm}^3/\text{min} and A_e from previous steps.

Explanation

1. Convert flow rate to cm³/min<br /> Given $R = 4.75 \, \text{lpm}$, convert to $\text{cm}^3/\text{min}$: $R = 4.75 \times 1000 = 4750 \, \text{cm}^3/\text{min}$.<br /><br />2. Calculate piston area<br /> Piston diameter $D_p = 6.25 \, \text{cm}$. Area $A_p = \pi \left(\frac{D_p}{2}\right)^2 = \pi \left(\frac{6.25}{2}\right)^2$.<br /><br />3. Calculate rod area<br /> Rod diameter $D_r = 4.75 \, \text{cm}$. Area $A_r = \pi \left(\frac{D_r}{2}\right)^2 = \pi \left(\frac{4.75}{2}\right)^2$.<br /><br />4. Calculate effective area for retraction<br /> Effective area $A_e = A_p - A_r$.<br /><br />5. Calculate retract speed<br /> **Formula**: $V = \frac{R}{A_e}$. Substitute $R = 4750 \, \text{cm}^3/\text{min}$ and $A_e$ from previous steps.
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